∫OA = 10, that is, A (10/0,0) and Ba ⊥ OA (X axis).

The abscissa of point B is 10.

∫B is on a straight line and y=4/5x.

∴ x = 10，y = 4/5× 10 = 8； B (10/0,8)

∫CB∑OA(x axis), B (10/0,8)

∴C(0,8)

⑵ The quadrilateral CDEB is a diamond; The reason for this is the following：

∫△dcb is folded into ⊿DEB along the straight line BD.

∴∠DBC=∠DBE,BC=BE

∫CB∑OA(x axis)

∴∠EDB=∠DBC=∠DBE

∴ED=EB=CB

∫CB∑DE，CB=DE

∴ Quadrilateral CDEB is a parallelogram

ED = EB

∴ parallelogram CDEB is a diamond.